5.85 g of NaCl are dissolved in 90 g of water. The mole fraction of NaCl is-

Answer

`0.0177`
`3.53 K m^(-1)`

Explanation

(a) Moles of KCI ` = (7.45)/(74.5) = 0.1`
(Molar mass of KCI =39+35.5 = 74.5)
Moles of water ` = (100)/(18) = 5.56`
Mole fractio of KCl `= (0.1)/(0.1+5.56) = 0.0177`
(b) `w_(B) = 9.0 g, w_(A) = 100 g, Delta T_(f) = 0.93^(@) , M_(B) = 342`
`Delta T_(f) = (K_(f) xx 1000 xx w_(B))/(w_(A) xx M_(B))`
or `K_(f) = (Delta T_(f) xx w_(A) xx M_(B))/(1000 xx w_(B)) = (0.93 xx 100 xx 342)/(1000 xx 9.0)`
` = 3.53 K m^(-1)`